Stats Exam 3 Estimation

Researchers from Dartmouth Medical School conducted a study in 2003 to look at the connection between watching actors smoking in movies and smoking initiation among adolescents. In the study, 6,522 U.S. adolescents ages 10-14 who had never tried smoking were randomly selected. Of those who subsequently tried smoking for the first time, 38% did so because of exposure to smoking in the movies. Estimate the proportion of all U.S. adolescents ages 10-14 who started smoking because of seeing actors smoke in movies by constructing a 95% confidence interval.

1. √.38(1−.38)/6522 = 0.006 2. .38 + 2(0.006) = 0.392 .38 - 2(0.006) = 0.368 so, (0.368, 0.392)

A researcher wanted to estimate µ, the mean number of hours that students at a large state university spend exercising per week. The researcher collects data from a sample of 150 students who leave the university gym following a workout. Which of the following is true regarding x̄, the average number of hours that the 150 sampled students exercise per week? A. It is an unbiased estimate for µ. B. It is not an unbiased estimate for µ and probably underestimates µ. C. It is not an unbiased estimate for µ and probably overestimates µ.

C. It is not an unbiased estimate for µ and probably overestimates µ. It is not an unbiased estimator for µ because the sample was not a random sample of 150 students from the entire student body. In addition, students who leave the university gym following a workout are likely students who exercise on a regular basis and therefore tend to exercise more, on average, than students in general.

A study was conducted to estimate μ, the mean commute distance that all employed U.S. adults travel to work. Suppose a random sample of 49 employed U.S. adults gives a mean commute distance of 22 miles and that from prior studies, the population standard deviation is assumed to be σ = 8.4 miles. Based on this information, what would be the point estimate for μ? a. 22 b. 1.2 c. 49 d. 8.4

a. 22 Correct. 22 is x̄, the sample mean commute distance which is the point estimator of μ, the population mean commute distance.

A study was done on pregnant women who smoke during their pregnancies. In particular, the researchers wanted to study the effect that smoking has on the pregnancy length. A sample 114 pregnant women who were smokers participated in the study and were followed until the birth of their child. At the end of the study, the collected data were analyzed and it was found that the average pregnancy length of the 114 women was 260 days. From a large body of research, it is known that the length of human pregnancy has a standard deviation of 16 days. In the previous activity, we calculated a 95% confidence interval for μ, the mean pregnancy length of women who smoke during their pregnancy based on the given information, and found it to be 260 +/- 3, or (257, 263). Assume now that the researcher wants to get a more precise interval estimation by reducing the margin of error from 3 to 2 while maintaining the same level of confidence. How many additional smoking pregnant women should the researcher sample? (Hint: calculate first what the total sample size must be in order to achieve this).

1. n = (2(bc of 95)*sd/2 (m)) ^2 2. n = 2*16/2 = 16 3. 16^2 = 256 4. 256 - 114 = 142 additional women

What sample size of U.S. adults do you need, if you would like to estimate the proportion of U.S. adults who are "pro-choice" with a 2.5% margin of error (at the 95% level)? If you were to use a random sample of size n = 640 U.S. adults (instead of what you found in question 1), what would the margin of error roughly be?

1/(.025)^2 = 1600 n=1/m2 so, √1/640 = 0.039 or 4%

A researcher would like to estimate p, the proportion of U.S. adults who support recognizing civil unions between gay or lesbian couples. Due to a limited budget, the researcher obtained opinions from a random sample of only 2,222 U.S. adults. With this sample size, the researcher can be 95% confident that the obtained sample proportion will differ from the true proportion (p) by no more than which of the following percentages (answers are rounded)? a. 0.04 b.0.75 c. 2.1 d. 3

1/√(2222) = 2.1

A study was done on pregnant women who smoked during their pregnancies. In particular, the researchers wanted to study the effect that smoking has on pregnancy length. A sample of 114 pregnant women who were smokers participated in the study and were followed until the birth of their child. At the end of the study, the collected data were analyzed and it was found that the average pregnancy length of the 114 women was 260 days. From a large body of research, it is known that length of human pregnancy has a standard deviation of 16 days. Based on this study, find a 95% confidence interval for μ, the mean pregnancy length of women who smoke during their pregnancy, and interpret your interval in context.

95% = 2 standard deviations 16/sqrt(114) = 1.5 260 + 2(1.5) = 263 260 - 2(1.5) = 257 so, 257, 263

In which of the following scenarios can we calculate a confidence interval for the population mean? Check all that apply. a. A random sample of 60 cell phone calls is selected and the mean length of calls is determined to be 3.25 minutes with a standard deviation of 4.2 minutes. b. A random sample of 15 women athletes is selected and their mean weight is determined to be 136 pounds. Female athlete weights are known to be normally distributed with a population standard deviation of 20 pounds. c. A random sample of 14 cell phone calls is selected and the mean length of calls is determined to be 3.25 minutes with a standard deviation of 4.2 minutes. d. A random sample of 35 women athletes is selected and their mean weight is determined to be 136 pounds. Female athlete weights are known to be normally distributed with a population standard deviation of 20 pounds.

a. A random sample of 60 cell phone calls is selected and the mean length of calls is determined to be 3.25 minutes with a standard deviation of 4.2 minutes. b. A random sample of 15 women athletes is selected and their mean weight is determined to be 136 pounds. Female athlete weights are known to be normally distributed with a population standard deviation of 20 pounds. d. A random sample of 35 women athletes is selected and their mean weight is determined to be 136 pounds. Female athlete weights are known to be normally distributed with a population standard deviation of 20 pounds. A confidence interval for the population mean can be calculated when the population is normally distributed, regardless of sample size, or when the sample size is sufficiently large (n > 30), regardless of population shape.

A medical researcher wanted to estimate μ, the mean weight of newborns born to women over the age of 40. The researcher chose a random sample of 100 pregnant women who were over 40, followed them through the pregnancy, and found that the mean weight of the 100 newborns was 3,035 grams. From past research, it is assumed that the weight of newborns has a standard deviation of σ = 500. The researcher calculated the 95% confidence interval for μ to be (2935, 3135). If the researcher wanted to maintain the 95% level of confidence but report a confidence interval with a smaller margin of error, which of the following will achieve that? a. Redo the study, but this time with a sample of size 64. b. Redo the study, but this time with a sample of size 225. c. Redo the study and choose a different sample of size 100.

b. Redo the study, but this time with a sample of size 225. A larger sample size reduces the margin of error (for a given level of confidence). In particular, by using a sample size of 225 instead of 100, the margin of error will be reduced from 100 to roughly 67.

A local fast food company sells a 16-oz soft drink. To determine if the soft drinks served contain less than the advertised 16 oz, a random sample of 20 drinks is chosen and the amount of drink in each cup is measured (without ice). The mean amount of soft drink was found to be 15.5 oz with a standard deviation of 0.4 oz. Based on the sample results we cannot claim that the soft drink contains less than 16 oz. Which form of statistical inference does this conclusion represent? a. point estimation b. interval estimation c. hypothesis testing

c. hypothesis testing We are testing a claim about the population (that the amount of drink is 16 oz) using the sample data as evidence against the claim. In this case, we were unable to reject the claim about the population. In hypothesis testing, we have some claim about the population, and we check whether or not the data obtained from the sample provide evidence against this claim.

T OR F in point estimation, as long as a sample is taken at random, the distribution of sample means is exactly centered at the value of population mean.

T

T OR F in point estimation, p an unbiased estimator for p.

T

When the sampling distribution of a statistic centers exactly around the parameter it estimates we can say that the statistic is which of the following? a. unbiased b. normally distributed c. statistically significant d. equal

a. unbiased

Based on survey results, the mean amount of credit card debt for U.S. adults under 35 is $8,724. This point estimate would be unbiased and most accurate if the survey were based on which of the following? a. A random sample of 1,500 U.S. adults under 35 b. A random sample of 2,800 U.S. adults under 35 c. A random sample of 3,000 U.S. college students d. A national television news station poll with 3,500 responses

b, A random sample of 2,800 U.S. adults under 35

in general what do we use as the point estimation for u, the population mean

x, the sample mean

T or F in point estimation, X is therefore said to be an unbiased estimator for μ

T

A survey of a random sample of 1,500 young Americans found that 87% had earned their high school diploma. Based on these results, the 95% confidence interval for the proportion of young Americans who have earned their high school diploma is (0.853,0.887) What is the margin of error for this confidence interval? a. 0.017 b. 0.95 c. 0.034 d. 0.87

a. 0.017 0.887 - 0.87

A study was conducted to estimate μ, the mean commute distance that all employed U.S. adults travel to work. Suppose a random sample of 49 employed U.S. adults gives a mean commute distance of 22 miles and that from prior studies, the population standard deviation is assumed to be σ = 8.4 miles. We are 95% confident that the mean commute distance to work of all employed U.S. adults falls between which of the following intervals? a. 19.6 to 24.4 b. 18.4 to 25.6 c. 20.8 to 23.2

a. 19.6 to 24.4

A random sample of 35 two-year colleges in 2008-2009 collected data on the in-state tuition costs and enrollment totals. Based on the sample, the researchers estimated the mean in-state tuition for all two-year colleges to be $2,380. Which form of statistical inference does this conclusion represent? a. point estimation b. hypothesis testing c. interval

a. point estimation

A researcher conducted a survey of graduating college students. 350 college students were randomly selected to answer the survey. One of the survey questions asked, "What are your plans after graduation?" 214 students said they were planning to find a job, 83 students said they were planning to attend graduate school, and 53 students did not reply to the question. What is the point estimate for the proportion of all college students who plan to find a job after graduation? a. 83 b. 0.24 c. 0.61 d. 214

c. 0.61 214/350

A random sample of 160 households is selected to estimate the mean amount spent on electric service. A 95% confidence interval was determined from the sample results to be ($151, $216). Which of the following is the correct interpretation of this interval? a. There is a 95% chance that the mean amount spent on electric service is between $151 and $216. b. We are 95% confident that the mean amount spent on electric service among the 160 households is between $151 and $216. c. We are 95% confident that the mean amount spent on electric service among all households is between $151 and $216. d. 95% of the households will have an electric bill between $151 and $216.

c. We are 95% confident that the mean amount spent on electric service among all households is between $151 and $216.

A random sample of 30 students taking statistics at a community college found the student's mean GPA to be 3.25. A 95% confidence interval for the mean GPA of all students taking statistics at this college was determined to be (3.07, 3.43). What is the margin of error for this confidence interval? a. 3.25 b. 0.36 c. 0.95 d. 0.18

d. 0.18 3.43-3.25

Assuming data come from a random sample, under which of the following conditions should we not calculate a confidence interval for a population mean? a. Population distribution is unknown and sample size is 50 individuals. b. Population is normally distributed and sample size is 50 individuals. c. Population is normally distributed and sample size is 20 individuals. d. Population distribution is unknown and sample size is 20 individuals.

d. Population distribution is unknown and sample size is 20 individuals.

Harris Interactive® conducted an online study from January 3 to 10, 2006 among a nationwide sample of 1,040 U.S. adults (ages 18 years and over), of whom 565 currently use organic foods in preparing meals. Figures for age by gender, race/ethnicity, education, region, and household income were weighted to reflect the total U.S. adult population. Propensity score weighting was used to adjust for respondents' propensity to be online. Construct a 99% confidence interval for the proportion of all U.S. adults who use organic foods in preparing meals.

mean = 565/1040 = .543 √.543(1−.543)/1040= 0.015 .543 + 3(0.015) .543 - 3(0.015)

Based on the poll's results, estimate p, the proportion of all U.S. adults who believe the use of marijuana should be legalized, with a 95% confidence interval. margin error?

√.56(1−.56)/1000 = 0.016 .56 + 2(0.016) = .59 .56 - 2(0.016) = .52 margin error = .59 -.56 = 0.03 or 3%

We say that a point estimator is unbiased if which of the following is true? a. Its sampling distribution is centered exactly at the parameter it estimates. b. Its value is always equal to the parameter it estimates. c. Its sampling distribution is normal. d. The standard deviation of its sampling distribution decreases as the sample size increases

a. Its sampling distribution is centered exactly at the parameter it estimates.

Suppose we take repeated random samples of 50 college students from the same population and determine a 95% confidence interval for the mean GPA from each sample. Which of the following statements is true regarding the confidence intervals? Check all that apply. a. The intervals are centered around the population mean GPA. b. The intervals are centered around the sample mean GPA. c. 95% of the intervals will contain the sample mean in the long run d. 95% of the intervals will contain the population mean in the long run.

b. The intervals are centered around the sample mean GPA. d. 95% of the intervals will contain the population mean in the long run.

A study was conducted to estimate μ, the mean number of weekly hours that U.S. adults use computers at home. Suppose a random sample of 81 U.S. adults gives a mean weekly computer usage time of 8.5 hours and that from prior studies, the population standard deviation is assumed to be σ = 3.6 hours. How large a sample of U.S. adults is needed in order to estimate μ with a 95% confidence interval of length 1.2 hours? a. 6 b. 12 c. 20 d. 144

d. 144 The sample size equation is n = (z*σ/m)2 where the margin of error (m) = length/2. We would like our confidence interval to be a 95% confidence interval (implying that z* = 2) and the confidence interval length should be 1.2, therefore the margin of error (m) = 1.2/2 = 0.6. The sample size we need in order to obtain this is: (z*σ/m)2 = ((2 * 3.6)/0.6)2 = 144.

A study on exercise habits used a random sample of 2,540 college students (1,220 females and 1,320 males). The study found the following: 818 of the females in the sample exercise on a regular basis. 924 of the males in the sample exercise on a regular basis. The average time that the 1742 students who exercise on a regular basis (818 + 924) spend exercising per week is 4.2 hours. 1. What is the point estimate for the proportion of all female college students who exercise on a regular basis? 2. What is the point estimate for the proportion of all college students who exercise on a regular basis? 3. Which of the following has a point estimate of 4.2 a. The mean time that all college students spend exercising per week b. The percentage of all college students who exercise on a regular basis c. The mean time that all college students who exercise on a regular basis spend exercising per week

1. 818/1220 = 0.67 2. 1742/2540 = 0.69 3. c. The mean time that all college students who exercise on a regular basis spend exercising per week

Based on survey results, the proportion of U.S. adults who own at least one Apple product (iphone, ipad, macbook, etc) is 0.42. This point estimate would be unbiased and most accurate if the survey were based on which of the following? a. A random sample of 1,000 U.S. adults b. A random sample of 1,800 U.S. adults c. A random sample of 2,500 U.S. college students

b. A random sample of 1,800 U.S. adults

A researcher conducted a survey of graduating college students. 350 college students were randomly selected to answer the survey. One of the survey questions asked, "How much do you owe in student loans?" The average student loan debt reported by respondents was $42,576 with a standard deviation of $2,801. $42,576 is the point estimate for which of the following? a. The proportion of all college students who graduate college in debt. b. The mean salary of all college students who graduate college c. The standard deviation of loan debt of all college students who graduate college. d. The mean loan debt of all college students who graduate college.

d. The mean loan debt of all college students who graduate college.

A researcher would like to estimate p, the proportion of U.S. adults who support raising the federal minimum wage. If the researcher would like to be 95% sure that the obtained sample proportion would be within 2.4% of p (the proportion in the entire population of U.S. adults), what sample size should be used? a. 6,945 b. 1737 c. 435 d. 42

b. 1737 1/m2 1/(.024)^2

The mean score on the quantitative reasoning part of the GRE (Graduate Record Examination) of non-U.S. citizens has an unknown mean, , and an assumed standard deviation =8. Based on a random sample of non-U.S. citizens who took the GRE in 2014 the 95% confidence interval for was calculated to be (153.6, 158.6). Suppose now that a different random sample of non-U.S. citizens who took the GRE in 2014 is chosen and that this sample is larger than the sample that produced the confidence interval above. Which of the following is the most likely 95% confidence interval for µ based on the larger sample? a. (153.4, 159.2) b. (153.9, 158.9) c. (153.8, 158.4) d. (153.5, 158.1)

d. (153.5, 158.1) When the sample size increases, the confidence interval gets narrower, and this confidence interval is narrower (width = 4.6) than the confidence interval that is based on the smaller sample size (width = 5).

A researcher would like to estimate p, the proportion of U.S. adults who support recognizing civil unions between gay or lesbian couples. If the researcher would like to be 95% sure that the obtained sample proportion would be within 1.5% of p (the proportion in the entire population of U.S. adults), what sample size should be used? a. 17,778 b. 4,445 c. 1,112

4,445 1/m2 1/(.015)^2

A study estimated that the mean number of children per family in the the United States is 1.3. This point estimate would be unbiased and most accurate if it were based on which of the following? A. A random sample of 10,000 U.S. families with children from the state of Utah B. A random sample of 500 U.S. families with children C. A random sample of 5,000 U.S. families with children D. A random sample of 1,000 U.S. families

D. A random sample of 1,000 U.S. families The estimate is based on a random sample (and is therefore unbiased) and is also based on a large sample, which makes it more accurate.

A study was conducted to estimate μ, the mean number of weekly hours that U.S. adults use computers at home. Suppose a random sample of 81 U.S. adults gives a mean weekly computer usage time of 8.5 hours and that from prior studies, the population standard deviation is assumed to be σ = 3.6 hours. The 95% confidence interval for the mean, μ, is (7.7, 9.3). Which of the following will provide a more informative (i.e., narrower) confidence interval than the 95% confidence interval? Check all that apply. a. Using a sample of size 400 (instead of 81) b. Using a sample of size 36 (instead of 81) c. Using a different sample of size 81 d. Using a 90% level of confidence (instead of 95%) e. Using a 99% level of confidence (instead of 95%)

a. Using a sample of size 400 (instead of 81) d. Using a 90% level of confidence (instead of 95%)